Well first of all its Cp - Cv = nR, where is your moles of gas, and its only true for an ideal gas. You find this about about halfway through the first semester of physical chemistry.
The formal definitions of Cv and Cp are
[tex] C_v = \frac{\partial U}{\partial T} [/tex] and
[tex] C_p = \frac{\partial H}{\partial T}[/tex]
Where U is the internal energy and H is the enthalpy, defined to be H = U + pV. But for an ideal gas, pV = nRT. Substitutiotn this into the defition for Cp we get
[tex] C_p = \frac{\partial (U + nRT)}{\partial T}[/tex]
Are you talking about [tex]\partial[/tex] ? Thats the symbol for a partial deriviative. Its like a deriviative but for functions of mroe than one variable. To calculate it, you treat the other variables as constants, except for the one youare differentiating against.
No, path has nothing to do with it. It follows directly from the definition of U, H, Cp and Cv, and our assumption of an ideal gas. If our gas it is not ideal, that equation does not hold.